## View Poll Results: Do you think the player should switch?

Voters
48. You may not vote on this poll
• Yes

20 41.67%
• No

13 27.08%
• Don't know

4 8.33%
• No difference

11 22.92%

# Thread: Problem on Mathematics - Probability

1. For a start, i would tend to agree that the objective probability is 1/3, 2/3 as mentioned in the above threads. However, my stand is that it does not make a difference whether he switches or not.

This is my explanation:

Let us look at the state space in actual: Given three boxes A, B, C.

There are two main events in general that can happen:

1) Player chooses the correct box prior -
Say the keys are in A and player (P) chooses A. Host (H) chooses B. The actions are illustrated as follows:

P = A, H = B ---- switch => lose

Similarly, P = A, H = B ---- switch => lose

If you account for all permutations for the above event of (1), the keys can be in box A, B or C and player chooses the right box. Hence, for (1), and six events in total in which the player loses if he switch.

2) Player chooses the box without the keys prior
Say the keys are in A and player chooses B. Host can only choose C. The actions are illustrated as follows:

P = B, H = C ---- switch => win

P = C, H = B ---- switch => win

For the three possible positions of the keys, six events in total in which the player wins if he switches .

The state space has 12 events in all possibility. Six in which he will lose if he switches, six in which he wins if he switches. Hence no nash equilibrium. There is no action that is superior to the other.

Given also the lack of nash equilibrium on the participants perspective, no baynesian equilibrium on the part of the host. He cannot exploit his informational superiority and lead the participant into losing.

Just my half cents worth... Dunno right or not

2. Originally posted by James_T
Couldn't resist giving my 2cents worth.

1/3 or 1/2 all depends on when you are computing.

If prior to the event, you are told that the host would open one of the 2 unpicked boxes and allow you to trade, then probability is 1/3 as explained very painstakingly by Roy.

If you fast-forward to the point where the host opens one box, and you are allowed to switch, the probability is 1/2 , since it is either this or that box.

Using conditional prob

P ( prize in A | prize not in C ) = P( prize in A and prize not in C) / P (prize not in C) = (1/3) / (2/3) = 1/2
Agree.

3. For those who are not convinced that switching is the better choice, go ahead and write a short program with a million trials. You'll find that 2 times out of 3, the prize is in the other box that the game show host did not open.

It is more convincing than arguing logic, because this puzzle is very counter-intuitive. It really seems like it should be 0.5-0.5 rather than 1/3-2/3.

4. I disagree here. Let's forget all the chim maths and logics for a moment. Imagine someone who came late and only joined the show after one box was opened. What does he see? Two boxes with one prize in one - 50-50 for a simple reason:
the probability to him and to the guy on stage must be the same.

I think the main flaw with guessing 1/3 and 2/3 is simply this:
there is a discontinuity at the point when a box is opened. We have two different situations before and after the box is open. There were 3 boxes before and 2 boxes after. The situations are not related except that you have the same host and player.

In real life though, it is possible that one box is more likely to contain the prize cos the host may try to trick the player (suggested somewhere in the thread already).

Just 2 cents. Someone writing a program??

5. Originally posted by toasty
For those who are not convinced that switching is the better choice, go ahead and write a short program with a million trials. You'll find that 2 times out of 3, the prize is in the other box that the game show host did not open.

It is more convincing than arguing logic, because this puzzle is very counter-intuitive. It really seems like it should be 0.5-0.5 rather than 1/3-2/3.
A program has to be written based an algorithm derived from proper logic in the first place. If the logic and assumptions are flawed, then the program results would also be similarly flawed.

All programs have to be thought out with logic before coding can be done.

The key question here, I believe, is whether probability should be applied to the initial choice at all.

6. Originally posted by Zerstorer
A program has to be written based an algorithm derived from proper logic in the first place. If the logic and assumptions are flawed, then the program results would also be similarly flawed.

All programs have to be thought out with logic before coding can be done.

The key question here, I believe, is whether probability should be applied to the initial choice at all.
The probabilities associated with the first choice is simply 1/3 for picking the right chest.

For example, you can set the choice to always be A out of A, B and C, and the location of the key is then random among A, B and C. So the Probability that the key is in A is 1/3.

Alternatively, you can consider both the choice as well as the location of the key as random among A, B and C. Then the probabiliy of amking the right choice is then =

P(choice = A and key in A) + P(choice = B and key in B) + P(choice = C and Key in C)
= 1/3 * 1/3 + 1/3 * 1/3 + 1/3 * 1/3 = 1/3

So probability of picking the right chest initially is 1/3 whether you randomize the choice and fix the key, radomise the key and fix the choice, or randomise both.

The question is still, after the host open the empty box and offers to trade, can that scenario be treated as entirely independent from the intitial choice, or dependent.

After the initial choice is made, the chest that the host would open is DEPENDENT on the initial choice. If the initial choice did not get the key, then the box that the host would open is entirely detemined, because out of the 2 chests not selected, 1 is emtpy and the other holds the key, and the host knows exactly which one to open.

If the initial choice hits the key, then the host has 2 choices, and he can randomly pick either one to open. This is on the condition that the initial pick is correct. So it is still dependent on the initial choice.

Since I have established that the choices of picking the right chest initialy is 1/3, let's say the key is in A.

So P(choice = A) = 1/3
P(choice <> A) = P(choice =B) + P(choice = C) = 1/3 + 1/3 = 2/3

case 1: P(host opens C | choice = A) = 1/2 * 1/3 = 1/6
case 2: P(host opens B | choice = A) = 1/2 * 1/3 = 1/6
case 3: P(host opens C | choice = B) = 1 * 1/3 - host has to open C if is the choice is B since the key is in A.
case 4: P(host opens B | choice = C) = 1 * 1/3 - host has to open B if the choice is C since the key is in A.

All other cases are not possible, such as P(host opens A | choice = A), by definition of the game procedure. So P(all other cases) = 0

So if the player switches, he will win in cases 3 and 4, so P(win if switches) = P(case 3) + P(case 4) = 2/3

If the player does not switch, he will win in cases 1 and 2, so P(win if does not swich) = P(case 1) + P(case 2) = 1/6 + 1/6 = 1/3.

Ergo, the probabilty of winning is the player swtiches is 2/3.

You can choose to do the same analysis by randomising both the key location and the initial choice. That would complicate the calculations but the results will be the same. The chances of winning by switching is 2/3.

Those who believe that the chances are 50-50 do so because you think the second choice (whether or not to switch) is equivalent to a random selection between 2 chests with the key in one of them.

If you think carefully this is not the case, as the 2 chests available in the second choice is DEPENDENT on the initial choice, and that the key did not change location randomly between the first and second choice. So the probabilities associated with the initial choice has to be carried over to the second choice.

Convinced yet?

7. Originally posted by hong2
I disagree here. Let's forget all the chim maths and logics for a moment. Imagine someone who came late and only joined the show after one box was opened. What does he see? Two boxes with one prize in one - 50-50 for a simple reason:
the probability to him and to the guy on stage must be the same.

I think the main flaw with guessing 1/3 and 2/3 is simply this:
there is a discontinuity at the point when a box is opened. We have two different situations before and after the box is open. There were 3 boxes before and 2 boxes after. The situations are not related except that you have the same host and player.

In real life though, it is possible that one box is more likely to contain the prize cos the host may try to trick the player (suggested somewhere in the thread already).

Just 2 cents. Someone writing a program??
The events happened whether or not the third person came late into the show to observe the events.

Just because the late comer chooses to ignore all prior events when considering the odds does not mean that that is correct and applicable to the player on stage.

Probability calculation is highly dependent on the events taken into consideration in the calculations. The less information taken into consideration, the less accurate the results are.

8. I have added one more option to the poll.

Option 1: you believe the odds of winning are higher if the player switches.
Option 2: you believe the odds of winning are higher if the player does not switch.
Option 3: you don't know.
Option 4: you believe the odds are the same whether or not the player switches.

I am guessing those who picked option 2 before I added option 4 would have picked option 4 if it were available. You can indicate by posting if you have made a vote and decides to jump ship.

9. How about changing the scenario a bit? Let's say the after contestant has chosen the box A, and then the host wants to create a sense of excitement to the audience, and he first opens box B and then box C WITHOUT allowing the contestant to rechoose.

Ok, when the host opened up box B, voila, it is empty. So the contestant will tell himself, "Okay my chance of winning is initially 33% but now it is 50% because Box B is empty."

It is similar to the case put up by Roygoh. The chance of winning for the contestant, in my view, is thus 50% after the host opened up the empty box. If the audience suspect that the host is on purpose not to let him win the grand prize, then he should switch. However, if the host is not playing any trick but just to add a twist to the game, then it is entirely up to the contestant to decide to switch or not to switch, cos the probability of winning is now 50-50.

10. Originally posted by roygoh
The events happened whether or not the third person came late into the show to observe the events.

Just because the late comer chooses to ignore all prior events when considering the odds does not mean that that is correct and applicable to the player on stage.

Probability calculation is highly dependent on the events taken into consideration in the calculations. The less information taken into consideration, the less accurate the results are.
Ermm... correct if I am wrong, we cannot use normal probability function to calculate the chance of the third person winning the game if we were to based on prior results. If prior results have to be considered, Bayesian theorem have to be applied. (Note: prior results is different from prior events.)

11. Originally posted by roygoh

Convinced yet?
Still no.

Still think that there is no need to apply Baye's Theorem in this case.

There isn't a need to take the initial choice into account given the latter scenario of only having to choose between 2.

Sometimes I feel that there shouldn't be a need to do math for the sake of math.

12. case 1: P(host opens C | choice = A) = 1/2 * 1/3 = 1/6
case 2: P(host opens B | choice = A) = 1/2 * 1/3 = 1/6
case 3: P(host opens C | choice = B) = 1 * 1/3 Convinced yet?
Not really

By the way, i think P(host opens C | choice = A) = P[(host opens C) and (choice = A)]/P( choice = A) = [1/2*1/3]/1/3 = 1/2.

Anyway, i choose i din know the last time cuz i believe its 50/50... So i'll add one to option 4.

13. Originally posted by Zerstorer
Still no.

Still think that there is no need to apply Baye's Theorem in this case.

There isn't a need to take the initial choice into account given the latter scenario of only having to choose between 2.

Sometimes I feel that there shouldn't be a need to do math for the sake of math.
I agree with Zerstorer that there isnt the need to take the initial choice into consideration, because they are mutually exclusive.

However, I do feel that at some point it is necessary to use Bayesian theorem.. hehe...

Anyway I asked the same question to one of the research fellows in NTU, and he also told me that the two events are mutually exclusive so the probability should be 50-50.

14. Originally posted by trendmatrix
I agree with Zerstorer that there isnt the need to take the initial choice into consideration, because they are mutually exclusive.
Oh..actually I think its because they are Independent events.

i.e P(A|B)=P(A)
P(B|A)=P(B)

The initial choice has totally no bearing on what goes on later.

Mutually exclusive events would mean that if A happens, B cannot happen.

15. Originally posted by Zerstorer
Oh..actually I think its because they are Independent events.

The initial choice has totally no bearing on what goes on later.

Mutually exclusive events would mean that if A happens, B cannot happen.
Oh sorry. I meant independent... getting old.. (in fact havent touched probability for a long long time...)

16. Originally posted by trendmatrix
Oh sorry. I meant independent... getting old.. (in fact havent touched probability for a long long time...)
Same here..in fact..lazy to work out all the permutations even..heh.

17. Now I see my mistake in my long post.

Let me revise:

Let say Key is in A, and the initial choice is random.

case 1: P(choice = B and host opens C) = P(choice = B) * P(host opens C | choice = B) = 1/3 * 1 = 1/3
case 2: p(choice = C and host opens B) = P(choice = C) * P(host opens C | choice = B) = 1/3 * 1 = 1/3
case 3: P(choice = A and host opens B) = P(choice = A) * P(host opens B | choice = A) = 1/3 * 1/2 = 1/6
case 4: P(choice = A and host opens C) = P(choice = A) * P(host opens C | choice = A) = 1/3 * 1/2 = 1/6

All other cases are not valid, by design of the game.

So if the player switches, he will win in case 1 and 2, so P(win if switch) = P(case 1) + P(case 2) = 2/3

If the player does not switch, he will win in case 3 and case 4, so P(win if not switch) = P(case 3) + P(case 4) = 1/3

So my position is still the same.

- Roy

18. How about looking at it this way.

Say the player already know how the host will conduct the game, and he makes a decision that he will switch no mater what.

So he goes on stage, makes a random choice, and switches when the host offers him the opportunity. His chances of winning is equal to his chances of not picking the correct chest initially. Thus 2/3.

Say the player decides that he will not switch no matter what prior to the game. His chances of winning is equal to his chances of picking the correct chest initially. Thus 1/3.

Does this sound like a valid argument to you?

- Roy

19. Originally posted by Zerstorer
The thing is that he is allowed to change his choice now. Whatever choice he made doesn't affect the choice he makes now. The opening of Box C is inconsequential, it will always be empty and it doesn't affect the decision in anyway.

Nope. Because the computation should be made at the branch when he is faced with only 2 boxes.

So if he isn't allowed to change, the probability remains 1/3.
Since he is allowed to change his choice, the decision actually occurs at the point when there is only 2 boxes left so its 1/2.
This is the type of puzzles created by mathematician to show that we are stupid and have pea size brain and they are superior being !
I agree with Zerstorer's logic.

Let's look at it from another way and and forget about the game show host.

Lets say you have chosen A and one of the other 2 boxes is opened and the key is in it ! So you are left with 2 boxes without key. Now, What is the porbability of either boxes having the key ? Still 1/3 ?

or

If you have 2 choices and you pick one and the other one is revealed to you. Now what is the probability of the one you choose being the correct one ? Still 50% or depends one what is in the one that was revealed to you ?

20. Originally posted by Gan CW
This is the type of puzzles created by mathematician to show that we are stupid and have pea size brain and they are superior being !
I agree with Zerstorer's logic.

Let's look at it from another way and and forget about the game show host.

Lets say you have chosen A and one of the other 2 boxes is opened and the key is in it ! So you are left with 2 boxes without key. Now, What is the porbability of either boxes having the key ? Still 1/3 ?
The box with the key will not be opened because the hosts knows which one has the key and he will never open that. This is a condition of the game that cannot be ignored. So it has to be taken into consideration. If you are saying that the host does not know where the key is and thus oipens one of the remaining boxes randomly, then the game rules have changed, and your statement is tru but does not apply to the original quesiton anymore.

Originally posted by Gan CW
or

If you have 2 choices and you pick one and the other one is revealed to you. Now what is the probability of the one you choose being the correct one ? Still 50% or depends one what is in the one that was revealed to you ? [/B]
I can see the reason behind this argument, but I don't think it applies because if there are only 2 choices, revealing the one not chosen is equivalent to reaveling the one chosen.

When there are 3 choices, revealing one of the 2 that were not chosen is not equivalent to revealing the one that was chosen because there is still one more unknown box. Also, the person choosing which one to reveal has total knowledge over where the key is and will never revel the one with the key.

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